T-SQL usando SUM para un total acumulado

No es necesario unirse, puede acelerar la consulta de esta manera:

select id, [user], value, total
from
(
  select id, [user], value, 
  row_number() over (partition by [user] order by id desc) rn, 
  sum(value) over (partition by [user]) total
from users
) a
where rn = 1